I have to find the density function of $|X|^{\frac{1}{2}}$ where $X$ follows a standard normal distribution. This is what I have attempted so far:
$$F_y(y)=P(Y\le y)=P(|X|^{\frac{1}{2}}\le y)=P(|X|\le y^2)=P(X\le y^2)=F_X(y^2)=\int^{y^2}_\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx.$$
I know that the density function $f_Y(y)=F_y'(y)$ but I don't know to deal with the integral.
First, there is a mistake in your derivation since $P(|X|\le y^2)=P(X\le y^2)$ is false. You have $$P(|X|\le y^2)=P(X\in[-y^2,y^2]) = \frac{1}{\sqrt{2\pi}}\int_{-y^2}^{y^2}e^{-x^2/2}dx\,.$$
Then your approach is correct, you have to derive the integral. How to do so? Notice that $$ \frac{1}{\sqrt{2\pi}}\int_{-y^2}^{y^2}e^{-x^2/2}dx = \frac{2}{\sqrt{2\pi}}\int_0^{y^2}e^{-x^2/2}dx$$ since the integrand is symmetric. To find the derivative you can see this integral as something in the form $F_Y(y) = G(y^2)$, where $G(y) = \frac{2}{\sqrt{2\pi}}\int_0^{y}e^{-x^2/2}dx$. So you just use the formula for the derivative of the composition of function. You have $$f_Y(y) = F_Y'(y) = 2y G'(y^2) = 2y\frac{2}{\sqrt{2\pi}}e^{-y^4/2}=\sqrt{\frac{8}{\pi}}ye^{-y^4/2}\,.$$